3311. Longest arithmetic

Ideas ： Enumerate each location , Set a starting point i Then find the last position of his arithmetic sequence j, We can find that at this time i——j The sequence length at any position in is less than i Start , Then we can go directly from j Position start enumeration

Code ：

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 2e5+5;
int w[N];
int main()
{
    
    int T;
    scanf("%d", &T);
    for(int cases = 1; cases <= T; cases ++){
    
        int n;
        
        scanf("%d", &n);
        for(int i = 0; i < n; i ++)
            scanf("%d", &w[i]);
            
        int res = 0;
        for(int i = 0; i < n; i ++){
    
            int j = i + 2;
            while(j < n && w[j] - w[j - 1] == w[j - 1] - w[j - 2]) j++;
            
            res = max(res, j -i);
            i = j - 2;
        }
        
        printf("Case #%d: %d\n", cases, res);
    }
    return 0;
}